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An electron emitted with zero velocity from the hot cathode in a vacuum tube is accelerated by the electric field towards the anode. If the anode is at positive potential of 1000 V with respect to the cathode, find the velocity acquired by the electron as it reaches the anode. (e = 1.6 x 10-19 C, m = 9.11 x 10-31 kg)?
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An electron emitted with zero velocity from the hot cathode in a vacuu...
Solution:
Given:
Potential difference, V = 1000 V
Charge of electron, e = 1.6 x 10^-19 C
Mass of electron, m = 9.11 x 10^-31 kg
We know that the potential energy gained by the electron is equal to the work done by the electric field on the electron.
Therefore, potential energy gained by the electron, PE = eV
This potential energy is converted into kinetic energy as the electron moves towards the anode.
Therefore, kinetic energy gained by the electron, KE = PE
So, (1/2)mv^2 = eV
Where v is the velocity acquired by the electron.
Solving for v, we get:
v = √(2eV/m)
Substituting the given values, we get:
v = √(2 x 1.6 x 10^-19 x 1000/9.11 x 10^-31)
v = 5.93 x 10^6 m/s
Therefore, the velocity acquired by the electron as it reaches the anode is 5.93 x 10^6 m/s.
Explanation:
When an electron is emitted from the cathode, it has an initial velocity of zero. However, when it enters the electric field between the cathode and the anode, it experiences a force due to the electric field. This force accelerates the electron towards the anode. The potential difference between the cathode and the anode determines the amount of work done by the electric field on the electron. This work done is converted into kinetic energy, which results in the electron acquiring a velocity. The velocity acquired by the electron can be calculated using the formula v = √(2eV/m), where V is the potential difference, e is the charge of the electron and m is the mass of the electron.
Given:
Potential difference, V = 1000 V
Charge of electron, e = 1.6 x 10^-19 C
Mass of electron, m = 9.11 x 10^-31 kg
We know that the potential energy gained by the electron is equal to the work done by the electric field on the electron.
Therefore, potential energy gained by the electron, PE = eV
This potential energy is converted into kinetic energy as the electron moves towards the anode.
Therefore, kinetic energy gained by the electron, KE = PE
So, (1/2)mv^2 = eV
Where v is the velocity acquired by the electron.
Solving for v, we get:
v = √(2eV/m)
Substituting the given values, we get:
v = √(2 x 1.6 x 10^-19 x 1000/9.11 x 10^-31)
v = 5.93 x 10^6 m/s
Therefore, the velocity acquired by the electron as it reaches the anode is 5.93 x 10^6 m/s.
Explanation:
When an electron is emitted from the cathode, it has an initial velocity of zero. However, when it enters the electric field between the cathode and the anode, it experiences a force due to the electric field. This force accelerates the electron towards the anode. The potential difference between the cathode and the anode determines the amount of work done by the electric field on the electron. This work done is converted into kinetic energy, which results in the electron acquiring a velocity. The velocity acquired by the electron can be calculated using the formula v = √(2eV/m), where V is the potential difference, e is the charge of the electron and m is the mass of the electron.
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An electron emitted with zero velocity from the hot cathode in a vacuum tube is accelerated by the electric field towards the anode. If the anode is at positive potential of 1000 V with respect to the cathode, find the velocity acquired by the electron as it reaches the anode. (e = 1.6 x 10-19 C, m = 9.11 x 10-31 kg)?
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An electron emitted with zero velocity from the hot cathode in a vacuum tube is accelerated by the electric field towards the anode. If the anode is at positive potential of 1000 V with respect to the cathode, find the velocity acquired by the electron as it reaches the anode. (e = 1.6 x 10-19 C, m = 9.11 x 10-31 kg)? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about An electron emitted with zero velocity from the hot cathode in a vacuum tube is accelerated by the electric field towards the anode. If the anode is at positive potential of 1000 V with respect to the cathode, find the velocity acquired by the electron as it reaches the anode. (e = 1.6 x 10-19 C, m = 9.11 x 10-31 kg)? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An electron emitted with zero velocity from the hot cathode in a vacuum tube is accelerated by the electric field towards the anode. If the anode is at positive potential of 1000 V with respect to the cathode, find the velocity acquired by the electron as it reaches the anode. (e = 1.6 x 10-19 C, m = 9.11 x 10-31 kg)?.
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